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Guest Message by DevFuse
 

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Hats on a death row!! One of my favorites puzzles!


Best Answer bonanova, 08 January 2008 - 06:25 PM

Spoiler for Here's my idea
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470 replies to this topic

#461 Finix

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Posted 15 November 2011 - 09:47 AM

I heard a similar one on khan academy
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#462 Morningstar

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Posted 22 November 2011 - 12:34 AM

Is this actually still being discussed? It was started at the end of 2007!
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The cake is a lie.

#463 zorrotornado

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Posted 23 November 2011 - 05:42 PM

easy one come on :P

Spoiler for please use spoilers


:)

Edited by plainglazed, 23 November 2011 - 11:42 PM.

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#464 bendigi

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Posted 09 December 2011 - 07:49 AM

I do have another solution skipping the math. Although, the odd/even solution is quite nice.

I would love to get some feedback on my solution:

The first guy that is being asked, simply tells the color of the man in front of him, if he is lucky he lives, if not, oh well he did it for the team.
Now, the next guy has already heard his color (black or red) and he is guaranteed freedom, but to help the next guy, here is my idea
If his color matches the one in front, he says the color without hesitation, if it's different, he pauses and thinks and then says his color.
The next guy listening, will consider the pause as a negation of the prevoius color and continue this process ... :)

Good one ?
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#465 Swinslow

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Posted 13 December 2011 - 12:38 PM

Your reasoning is correct. It was stated that a correct strategy would guarantee the safety of 19 of the 20, with the 20th (first to guess) having a 50% chance.

With no other information present, I don't see how this is solvable. What strategy would allow you to say the word "red" or the word "black" and tell nineteen people which of two colors each was wearing? I believe it is impossible; you cannot convey that much information in a simple binary choice.

I can see a method to save at least 13 of the 20 people:

Spoiler for solution


I see but knowing prisoners.....

Spoiler for

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#466 Tyler Durden

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Posted 03 January 2012 - 06:46 PM

Sorry. I must have missed something in the instructions. This solution only works when there are 10 black hats and 10 red hats. No where in the instructions does it say this. The solution fails if there are different numbers of black and red hats. There could be 19 black hats and 1 red hat.
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#467 Chew Jing Kai

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Posted 04 January 2012 - 03:23 PM

All 20 prisoners will be freed

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20

Prisoner 20 will start first. Since he can see everyone in front of him, he knows and will shout the color of his hat (if he sees 9 black 10 red that means he is wearing black, and vice versa)
Prisoner 19 will also be able to shout the color of his hat (if he sees 9 black 9 red and prisoner 20 shouted black, that means he is wearing red)
Prisoners 18 to 2 repeat the same steps.
Prisoner 1 having heard all the color shouted before his turn will also know his color (if 9 shouted black and 10 shouted red that means he is black)
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#468 Ranbir Singh

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Posted 05 January 2012 - 03:11 PM

Here's my solution, not sure if it's allowed or not...

Spoiler for Here's my solution.. probably something wrong with it...


How's this one?

one wrong timing and all are gone !!
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#469 Brainiac101

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Posted 07 January 2012 - 02:16 AM

1.If the last person is right you say the opposite color if he is wrong you say the same color
2.You all say the same color and hope your right
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#470 vivekkumarjha

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Posted 13 June 2012 - 05:38 PM

well, nice one
but the slightest mistake and all would be executed

Edited by vivekkumarjha, 13 June 2012 - 05:39 PM.

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